Exercise 3.3 - Chapter 3 Theory of Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.3$
Question 1.
Solve the cubic equation: $2 x^3-x^2-18 x+9=0$ if sum of two of its roots vanishes.
Solution:
The given equation is $2 x^3-x^2-18 x+9=0$
$x^3-\frac{x^2}{2}-9 x+\frac{9}{2}=0$
Let the roots be $\alpha,-\alpha, \beta$
$
\begin{aligned}
& \alpha-\alpha+\beta=-\left(\frac{-1}{2}\right) \\
& \Rightarrow \beta=\frac{1}{2}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow-\alpha^2\left(\frac{1}{2}\right)=\frac{-9}{2} \\
& \alpha^2=9 \\
& \alpha=\pm 3
\end{aligned}
$
The roots are $3,-3, \frac{1}{2}$
Question 2.
Solve the equation $9 x^3-36 x^2+44 x-16=0$ if the roots form an arithmetic progression.
Solution:
The given equation is $9 \mathrm{x}^3-36 \mathrm{x}^2+44 \mathrm{x}-16=0$
$
x^3-4 x^2+\frac{44}{9} x-\frac{16}{9}=0
$
Let the roots be $\alpha-\mathrm{d}, \alpha, \alpha+\mathrm{d}$
As they are in A.P
$
\begin{aligned}
\alpha-d+\alpha+\alpha+d & =-(-4) & & \Rightarrow 3 \alpha=4 \quad \Rightarrow \alpha=\frac{4}{3} \\
(\alpha-d)(\alpha)(\alpha+d) & =-\left(\frac{-16}{9}\right) & & \Rightarrow \alpha\left(\alpha^2-d^2\right)=\frac{16}{9} \\
\frac{4}{3}\left(\frac{16}{9}-d^2\right) & =\frac{16}{9} & & \Rightarrow \frac{16}{9}-d^2=\frac{16}{9} \times \frac{3}{4} \\
\frac{16}{9}-\frac{4}{3} & =d^2 & & \Rightarrow d=\pm \frac{2}{3}
\end{aligned}
$

The roots are $\alpha-d, \alpha, \alpha+d$, when $d=\pm \frac{2}{3}, \alpha=\frac{4}{3}$
(i) $d=\frac{2}{3}, \alpha=\frac{4}{3}$
$
\begin{array}{c|c|c}
\alpha-d, & \alpha, & \alpha+d \\
\frac{4}{3}-\frac{2}{3}=\frac{2}{3} & \frac{4}{3} & \frac{4}{3}+\frac{2}{3}=\frac{6}{3}=2 \\
\alpha-d, & \alpha, & \alpha+d \\
\frac{4}{3}-\left(\frac{-2}{3}\right)=\frac{6}{3}=2 & \frac{4}{3} & \frac{4}{3}-\frac{2}{3}=\frac{2}{3}
\end{array}
$
(ii) $d=\frac{-2}{3}, \alpha=\frac{4}{3}$
Question 3.

Solve the equation $3 x^3-26 x^2+52 x-24=0$ if its roots form a geometric progression.
Solution:
The given equation is $3 x^3-26 x^2+52 x-24=0$
$
x^3-\frac{26}{3} x^2+\frac{52}{3} x-8=0
$

Given that the root are GP
$\therefore$ The roots are $\frac{\alpha}{r}, \alpha, \alpha r$
$
\begin{aligned}
& \frac{\alpha}{r} \times \alpha \times \alpha r=-(-8) \\
& \alpha^3=8 \quad \Rightarrow \alpha=2 \\
& \frac{\alpha}{r}+\alpha+\alpha r=-\left(\frac{-26}{3}\right) \quad \Rightarrow \alpha\left(\frac{1}{r}+1+r\right)=\frac{26}{3} \\
& 2\left(\frac{1}{r}+1+r\right)=\frac{26}{3} \\
& 3\left(\frac{1+r+r^2}{r}\right)=13 \quad \Rightarrow 3+3 r+3 r^2=13 r \\
& 3 r^2-10 r+3=0 \quad \Rightarrow(3 r-1)+(r-3)=0 \\
& r=\frac{1}{3}, r=3 \\
&
\end{aligned}
$
(i) $\alpha=2, r=3 \quad \Rightarrow$ The roots are $\frac{2}{3}, 2,6$
(ii) $\alpha=2, r=\frac{1}{3} \quad \Rightarrow$ The roots are $6,2, \frac{2}{3}$
Question 4.
Determine $f t$ and solve the equation $2 x^3-6 x^2+3 x+k=0$ if one of its roots is twice the sum of the other two roots.
Solution:
The given equation is $2 \mathrm{x}^3-6 \mathrm{x}^2+3 \mathrm{x}+\mathrm{k}=0$

$
\div 2 \Rightarrow x^3-3 x^2+\frac{3}{2} x+\frac{k}{2}=0
$
Given that
$
\begin{aligned}
\alpha+\beta+\gamma & =-(-3) & & \\
\frac{3 \alpha}{2} & =3 & & \\
\alpha & =2 & & \Rightarrow \beta+\gamma=1 \\
\alpha \beta+\beta \gamma+\gamma \alpha & =\frac{3}{2} & & \Rightarrow 2(\beta+\gamma)+\beta \gamma=\frac{3}{2} \\
2 \beta+\beta \gamma+2 \gamma & =\frac{3}{2} & & \Rightarrow \beta \gamma=\frac{3}{2}-2=\frac{-1}{2} \\
2(1)+\beta \gamma & =\frac{3}{2} & & \\
\beta \gamma & =\frac{-1}{2} & & \Rightarrow k=+2 \\
\alpha \beta \gamma & =-\frac{k}{2} & &
\end{aligned}
$
(1) $\Rightarrow \beta=1-\gamma$ substitute in (2)
$
\begin{array}{rlrl}
\gamma(1-\gamma)=\frac{-1}{2} & \Rightarrow \gamma-\gamma^2=\frac{-1}{2} \\
2 \gamma-2 \gamma^2=-1 & \Rightarrow 2 \gamma^2-2 \gamma-1=0 \\
\gamma & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{2 \pm \sqrt{4-4(2)(-1)}}{2(2)}=\frac{2 \pm \sqrt{4+8}}{4} \\
\gamma & =\frac{2 \pm 2 \sqrt{3}}{4}=\frac{1 \pm \sqrt{3}}{2} &
\end{array}
$
The roots are $2, \frac{1 \pm \sqrt{3}}{2}$ and $k=2$

Question 5.
Find all zeros of the polynomial $x^6-3 x^5-5 x^4+22 x^3-39 x^2-39 x+135$, if it is known that $1+2 i$ and $\sqrt{3}$ are two of its zeros.
Solution:
The given equation is $x^6-3 x^5-5 x^4+22 x^3-39 x^2-39 x+135=0$
The given roots are $1+2 i, \sqrt{3}$
The other roots are $1-2 i,-\sqrt{3}$
The factors are
$
\begin{aligned}
& =\left\{x^2-x(2)+(1+4)\right\}\{(x+\sqrt{3})(x-\sqrt{3})\} \\
& =\left(x^2-2 x+5\right)\left(x^2-3\right) \\
& =x^4-3 x^2-2 x^3+6 x+5 x^2-15 \\
& =x^4-2 x^3+2 x^2+6 x-15
\end{aligned}
$
To find this roots,

The other factor is $x^2-x-9=0$
$\therefore$ The roots are $1 \pm 2 i, \pm \sqrt{3}, \frac{1 \pm \sqrt{37}}{2}$
Question 6.
Solve the cubic equations:
(i) $2 \mathrm{x}^3-9 \mathrm{x}^2+10 \mathrm{x}=3$
(ii) $8 x^3-2 x^2-7 x+3=0$
Solution:
(i) Given equation is $2 x^3-9 x^2+10 x=3$
Sum of the co-efficients $=0$ $(\mathrm{x}-1)$ is a factor.
The other factor is $2 \mathrm{x}^2-7 \mathrm{x}+3$
$
\begin{aligned}
& 2 x^2-7 x+3=0 \\
& (x-3)(2 x-1)=0
\end{aligned}
$
The roots are $1,3, \frac{1}{2}$

(ii) Given equation is $8 x^3-2 x^2-7 x+3=0$
Sum of odd co-efficients $=$ Sum of even co-efficients $(\mathrm{x}+1)$ is a factor.
The other factor is $8 x^2-10 x+3$
$8 x^2-10 x+3=0$
$(4 \mathrm{x}-3)(2 \mathrm{x}-1)=0$
The roots are $\frac{3}{4}, \frac{1}{2},-1$

Question 7.
Solve the equation: $x^4-14 x^2+45=0$.
Solution:
The given equation is $\mathrm{x}^4-14 \mathrm{x}^2+45=0$
Let $x^2=y$
$
\begin{aligned}
& y^2-14 y+45=0 \\
& (y-9)(y-5)=0 \\
& y=9,5 \\
& x^2=9, x^2=5 \\
& x=\pm 3, x=\pm \sqrt{5}
\end{aligned}
$
The roots are $\pm 3, \pm \sqrt{5}$